POJ 3468 A Simple Problem with Integers（线段树区间求和）

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Description

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a₊₁, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a₊₁, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

Hint

The sums may exceed the range of 32-bit integers.

区间求和：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>typedef long long LL;using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )const int maxn=100000;int num[maxn];LL sum[maxn<<2],add[maxn<<2];int N,Q;void pushup(int rs){    sum[rs]=sum[rs<<1]+sum[rs<<1|1];}void pushdown(int rs,int l){    if(add[rs])    {        add[rs<<1]+=add[rs];        add[rs<<1|1]+=add[rs];        sum[rs<<1]+=add[rs]*(l-(l>>1));        sum[rs<<1|1]+=add[rs]*(l>>1);        add[rs]=0;    }}void build(int rs,int l,int r){     if(l==r)     {         scanf("%I64d",&sum[rs]);         return ;     }     int mid=(l+r)>>1;     build(rs<<1,l,mid);     build(rs<<1|1,mid+1,r);     pushup(rs);}void update(int c,int x,int y,int l,int r,int rs){     if(l>=x&&r<=y)     {         add[rs]+=c;         sum[rs]+=(LL)c*(r-l+1);         return ;     }     pushdown(rs,r-l+1);     int mid=(l+r)>>1;     if(x<=mid)   update(c,x,y,l,mid,rs<<1);     if(y>mid)    update(c,x,y,mid+1,r,rs<<1|1);     pushup(rs);}LL query(int x,int y,int l,int r,int rs){    if(l>=x&&r<=y)        return  sum[rs];    pushdown(rs,r-l+1);    int mid=(l+r)>>1;    LL ans=0;    if(x<=mid)   ans+=query(x,y,l,mid,rs<<1);    if(y>mid)    ans+=query(x,y,mid+1,r,rs<<1|1);    return ans;}int main(){     int x,y,z;     std::ios::sync_with_stdio(false);     while(~scanf("%d%d",&N,&Q))     {         CLEAR(sum,0);         CLEAR(add,0);         build(1,1,N);         char str[2];         while(Q--)         {             scanf("%s",str);             if(str[0]=='C')             {                 scanf("%d%d%d",&x,&y,&z);                 update(z,x,y,1,N,1);             }             else             {                 scanf("%d%d",&x,&y);                 printf("%I64d\n",query(x,y,1,N,1));             }         }     }     return 0;}

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